Question

Solve : \( \frac{dy}{dx} = \frac{x + 2y - 3}{2x + y - 3} \).

Hint:

To quickly check if a DE of the form \( \frac{dy}{dx} = \frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2} \) is reducible to homogeneous form, check the ratio of coefficients. If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the method of shifting the origin to the intersection point will work. If the ratios are equal, the equation is solved using the substitution \( z = a_1x + b_1y \).

Answer 1

Step 1: Understanding the Concept:
This is a non-homogeneous first-order differential equation of the form \( \frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2} \). Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) (\(\frac{1}{2} \neq \frac{2}{1}\)), the lines intersect. This type of equation can be reduced to a homogeneous differential equation by shifting the origin to the point of intersection of the lines \( x+2y-3=0 \) and \( 2x+y-3=0 \).
Step 2: Key Formula or Approach:
1. Find the point of intersection (h, k) of the two lines in the numerator and denominator.
2. Substitute \( x = X + h \) and \( y = Y + k \). This transforms the equation into a homogeneous form \( \frac{dY}{dX} = f(\frac{Y}{X}) \).
3. Solve the homogeneous equation using the substitution \( Y = vX \).
4. Separate variables, integrate, and substitute back to get the solution in terms of x and y.
Step 3: Detailed Explanation:
First, find the intersection of the lines \( x+2y-3=0 \) and \( 2x+y-3=0 \).
Solving these simultaneously: \[ x + 2y = 3 \quad \text{(1)} \] \[ 2x + y = 3 \quad \text{(2)} \] Multiplying (2) by 2 gives \( 4x + 2y = 6 \). Subtracting (1) from this gives \( 3x = 3 \Rightarrow x = 1 \).
Substituting \( x = 1 \) into (2) gives \( 2(1) + y = 3 \Rightarrow y = 1 \).
The point of intersection is (h, k) = (1, 1).
Now, let \( x = X + 1 \) and \( y = Y + 1 \). Then \( dx = dX \) and \( dy = dY \), so \( \frac{dy}{dx} = \frac{dY}{dX} \).
Substituting into the original equation: \[ \frac{dY}{dX} = \frac{(X+1) + 2(Y+1) - 3}{2(X+1) + (Y+1) - 3} = \frac{X + 2Y}{2X + Y} \] This is a homogeneous equation. Let \( Y = vX \). Then \( \frac{dY}{dX} = v + X\frac{dv}{dX} \). \[ v + X\frac{dv}{dX} = \frac{X + 2vX}{2X + vX} = \frac{1 + 2v}{2 + v} \] \[ X\frac{dv}{dX} = \frac{1 + 2v}{2 + v} - v = \frac{1 + 2v - v(2+v)}{2+v} = \frac{1 - v^2}{2+v} \] Separate the variables: \[ \frac{2+v}{1-v^2} dv = \frac{dX}{X} \] Using partial fractions for the left side: \( \frac{2+v}{(1-v)(1+v)} = \frac{A}{1-v} + \frac{B}{1+v} \).
This gives \( 2+v = A(1+v) + B(1-v) \). For \( v=1 \), \( 3 = 2A \Rightarrow A = 3/2 \). For \( v=-1 \), \( 1 = 2B \Rightarrow B = 1/2 \).
So the equation becomes: \[ \int \left( \frac{3/2}{1-v} + \frac{1/2}{1+v} \right) dv = \int \frac{dX}{X} \] Integrating both sides: \[ -\frac{3}{2}\ln|1-v| + \frac{1}{2}\ln|1+v| = \ln|X| + C_1 \] \[ \ln|1+v| - 3\ln|1-v| = 2\ln|X| + 2C_1 \] \[ \ln\left|\frac{1+v}{(1-v)^3}\right| = \ln(X^2) + C_2 \] \[ \frac{1+v}{(1-v)^3} = e^{C_2} X^2 = C X^2 \] Substitute back \( v = Y/X \): \[ \frac{1 + Y/X}{(1 - Y/X)^3} = C X^2 \Rightarrow \frac{X(X+Y)}{X^3} \cdot \frac{X^3}{(X-Y)^3} = C X^2 \Rightarrow \frac{X+Y}{(X-Y)^3} = C \] Finally, substitute back \( X = x-1 \) and \( Y = y-1 \): \[ \frac{(x-1)+(y-1)}{((x-1)-(y-1))^3} = C \Rightarrow \frac{x+y-2}{(x-y)^3} = C \] \[ x+y-2 = C(x-y)^3 \] Step 4: Final Answer:
The general solution to the differential equation is \( x + y - 2 = C(x - y)^3 \).